题解 | 反序数

利用函数的解法：#include<stdio.h> using namespace std; //设N是一个四位数，它的9倍恰好是其反序数（例如：1234的反序数是4321）求N的值 int Reverse(int n) { int reverse = 0; int remain; while (true) { remain = n % 10; n = n / 10; reverse = reverse * 10 + remain; if (n == 0) { break; } } return reverse; } int main() { //int n = 1234; //int reverse = 0; //int remain; ////while (n > 0) { //// remain = n % 10; //// n = n / 10; //// reverse = reverse * 10 + remain; ////} //while (true) {//纯C写while(1)即可 // remain = n % 10; // n = n / 10; // reverse = reverse * 10 + remain; // //死循环方式 // if (n == 0) { // break; // } //} //1000~9999 /*int n = 1234; int reverse = Reverse(n);*/ int n;//注意n的位置对程序的影响 for (n = 1000; n <= 9999; ++n) { if (n * 9 == Reverse(n)) { printf("%d\n", n); } } return 0; }</stdio.h>